How to prove that {$sin(x) , sin(2x) , sin(3x) ,…,sin(nx)$} is independent in $mathbb{R}$? [duplicate]
This question already has an answer here:
How to prove that the set ${sin(x),sin(2x),…,sin(mx)}$ is linearly independent? [on hold]
5 answers
Prove that {$sin(x) , sin(2x) , sin(3x) ,...,sin(nx)$} is independent in $mathbb{R}$
my trial :
we know that the Wronsekian shouldn't be $0$ to get the trivial solution and thus they are independent. its not trivial to show that $ W not = 0$
W =
begin{vmatrix}
(1)sin(x) & (1)sin(2x) & (1)sin(3x) & ... & (1)sin(nx) \
(1)cos(x) & (2)cos(2x) & (3)cos(3x) & ... & (n)cos(nx) \
-(1)^2sin(x) & -(2)^2sin(2x) & -(3)^2sin(3x) & ... & -(n)^2sin(nx) \
-(1)^3cos(x) & -(2)^3cos(2x) & -(3)^3cos(3x) & ... & -(n)^3cos(nx) \
end{vmatrix}
and so on. it looks like Vandermonde matrix but i cant prove that and so we conclude that its $Wnot =0$
differential-equations
marked as duplicate by Martin R, TheSimpliFire, 5xum, Shubham Johri, Ahmad Bazzi 16 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 5 more comments
This question already has an answer here:
How to prove that the set ${sin(x),sin(2x),…,sin(mx)}$ is linearly independent? [on hold]
5 answers
Prove that {$sin(x) , sin(2x) , sin(3x) ,...,sin(nx)$} is independent in $mathbb{R}$
my trial :
we know that the Wronsekian shouldn't be $0$ to get the trivial solution and thus they are independent. its not trivial to show that $ W not = 0$
W =
begin{vmatrix}
(1)sin(x) & (1)sin(2x) & (1)sin(3x) & ... & (1)sin(nx) \
(1)cos(x) & (2)cos(2x) & (3)cos(3x) & ... & (n)cos(nx) \
-(1)^2sin(x) & -(2)^2sin(2x) & -(3)^2sin(3x) & ... & -(n)^2sin(nx) \
-(1)^3cos(x) & -(2)^3cos(2x) & -(3)^3cos(3x) & ... & -(n)^3cos(nx) \
end{vmatrix}
and so on. it looks like Vandermonde matrix but i cant prove that and so we conclude that its $Wnot =0$
differential-equations
marked as duplicate by Martin R, TheSimpliFire, 5xum, Shubham Johri, Ahmad Bazzi 16 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Note that $LaTeX$ commands $sin$sin
and $cos$cos
for future reference.
– TheSimpliFire
16 hours ago
thanks i updated it
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
1
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago
|
show 5 more comments
This question already has an answer here:
How to prove that the set ${sin(x),sin(2x),…,sin(mx)}$ is linearly independent? [on hold]
5 answers
Prove that {$sin(x) , sin(2x) , sin(3x) ,...,sin(nx)$} is independent in $mathbb{R}$
my trial :
we know that the Wronsekian shouldn't be $0$ to get the trivial solution and thus they are independent. its not trivial to show that $ W not = 0$
W =
begin{vmatrix}
(1)sin(x) & (1)sin(2x) & (1)sin(3x) & ... & (1)sin(nx) \
(1)cos(x) & (2)cos(2x) & (3)cos(3x) & ... & (n)cos(nx) \
-(1)^2sin(x) & -(2)^2sin(2x) & -(3)^2sin(3x) & ... & -(n)^2sin(nx) \
-(1)^3cos(x) & -(2)^3cos(2x) & -(3)^3cos(3x) & ... & -(n)^3cos(nx) \
end{vmatrix}
and so on. it looks like Vandermonde matrix but i cant prove that and so we conclude that its $Wnot =0$
differential-equations
This question already has an answer here:
How to prove that the set ${sin(x),sin(2x),…,sin(mx)}$ is linearly independent? [on hold]
5 answers
Prove that {$sin(x) , sin(2x) , sin(3x) ,...,sin(nx)$} is independent in $mathbb{R}$
my trial :
we know that the Wronsekian shouldn't be $0$ to get the trivial solution and thus they are independent. its not trivial to show that $ W not = 0$
W =
begin{vmatrix}
(1)sin(x) & (1)sin(2x) & (1)sin(3x) & ... & (1)sin(nx) \
(1)cos(x) & (2)cos(2x) & (3)cos(3x) & ... & (n)cos(nx) \
-(1)^2sin(x) & -(2)^2sin(2x) & -(3)^2sin(3x) & ... & -(n)^2sin(nx) \
-(1)^3cos(x) & -(2)^3cos(2x) & -(3)^3cos(3x) & ... & -(n)^3cos(nx) \
end{vmatrix}
and so on. it looks like Vandermonde matrix but i cant prove that and so we conclude that its $Wnot =0$
This question already has an answer here:
How to prove that the set ${sin(x),sin(2x),…,sin(mx)}$ is linearly independent? [on hold]
5 answers
differential-equations
differential-equations
edited 14 hours ago
wilsonw
467315
467315
asked 16 hours ago
Mather
1297
1297
marked as duplicate by Martin R, TheSimpliFire, 5xum, Shubham Johri, Ahmad Bazzi 16 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, TheSimpliFire, 5xum, Shubham Johri, Ahmad Bazzi 16 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Note that $LaTeX$ commands $sin$sin
and $cos$cos
for future reference.
– TheSimpliFire
16 hours ago
thanks i updated it
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
1
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago
|
show 5 more comments
Note that $LaTeX$ commands $sin$sin
and $cos$cos
for future reference.
– TheSimpliFire
16 hours ago
thanks i updated it
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
1
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago
Note that $LaTeX$ commands $sin$
sin
and $cos$ cos
for future reference.– TheSimpliFire
16 hours ago
Note that $LaTeX$ commands $sin$
sin
and $cos$ cos
for future reference.– TheSimpliFire
16 hours ago
thanks i updated it
– Mather
16 hours ago
thanks i updated it
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
1
1
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago
|
show 5 more comments
1 Answer
1
active
oldest
votes
If $sum_{j=1}^{n} c_j sin (jx)$=0 for all $x$ you can prove that each $c_k=0$ by multiplying both sides by $sin(kx)$ and and integrating from $-pi$ to $pi$. Use the fact that $int_{-pi}^{pi} sin(jx)sin (kx), dx =0$ if $j neq k$. This proves the stronger result that the sequence is independent on $[-pi,pi]$.
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $sum_{j=1}^{n} c_j sin (jx)$=0 for all $x$ you can prove that each $c_k=0$ by multiplying both sides by $sin(kx)$ and and integrating from $-pi$ to $pi$. Use the fact that $int_{-pi}^{pi} sin(jx)sin (kx), dx =0$ if $j neq k$. This proves the stronger result that the sequence is independent on $[-pi,pi]$.
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
add a comment |
If $sum_{j=1}^{n} c_j sin (jx)$=0 for all $x$ you can prove that each $c_k=0$ by multiplying both sides by $sin(kx)$ and and integrating from $-pi$ to $pi$. Use the fact that $int_{-pi}^{pi} sin(jx)sin (kx), dx =0$ if $j neq k$. This proves the stronger result that the sequence is independent on $[-pi,pi]$.
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
add a comment |
If $sum_{j=1}^{n} c_j sin (jx)$=0 for all $x$ you can prove that each $c_k=0$ by multiplying both sides by $sin(kx)$ and and integrating from $-pi$ to $pi$. Use the fact that $int_{-pi}^{pi} sin(jx)sin (kx), dx =0$ if $j neq k$. This proves the stronger result that the sequence is independent on $[-pi,pi]$.
If $sum_{j=1}^{n} c_j sin (jx)$=0 for all $x$ you can prove that each $c_k=0$ by multiplying both sides by $sin(kx)$ and and integrating from $-pi$ to $pi$. Use the fact that $int_{-pi}^{pi} sin(jx)sin (kx), dx =0$ if $j neq k$. This proves the stronger result that the sequence is independent on $[-pi,pi]$.
edited 15 hours ago
answered 16 hours ago
Kavi Rama Murthy
50.7k31854
50.7k31854
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
add a comment |
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
This question has been asked and answered before ...
– Martin R
16 hours ago
This question has been asked and answered before ...
– Martin R
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
thank you but is there an answer related to ODE and wronsekian
– Mather
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
@Mather please emphasize in your question that you would like a Wronsekian oriented answer.
– Ahmad Bazzi
16 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
the solution is simple as they mentioned above , yet the integral above proves that they are independent only in $ [-pi,pi] $ right ? can we do that for $Re$ in a similar way ?
– Mather
15 hours ago
1
1
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
@Mather Independent on a subset of $mathbb R$ implies indepedence on all of $mathbb R$.
– Kavi Rama Murthy
15 hours ago
add a comment |
Note that $LaTeX$ commands $sin$
sin
and $cos$cos
for future reference.– TheSimpliFire
16 hours ago
thanks i updated it
– Mather
16 hours ago
this is ODE class i don't know whats the topic there but it doesn't like ODE class
– Mather
16 hours ago
@MartinR And of course that one deserves closure as well
– TheSimpliFire
16 hours ago
1
@Mather The answer I have given does not use any theorem. It is extremely elementary.
– Kavi Rama Murthy
16 hours ago